# Replacing symbols with numbers in a system of linear equations

Posted 7 months ago
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 I'm trying to simulate a physical problem through a system of linear equations in Mathematica. The system has a complete symbolic answer when the dimensions of the device are not evaluated but when I add them to the solution, the system finds the answers relative to each other and it seems to need three more equations to give independent answers. What could be the cause of the system not having an answer when parameters are replaced with numbers?Also in a specific case where the initial condition allows the answers to be independent when the dimensions and initial conditions are added before or after the solution of the system, the answer differs. Which answer should i assume to be the correct one? Here is my system of equations: F1x + F2x == 0 F1y + F2y == 0 -F1y L0 + F2y L0 + T1z + T2z + d F1x Cos[\[Theta]1] + d F2x Cos[\[Theta]2] + d F1y Sin[\[Theta]1] + d F2y Sin[\[Theta]2] - F1y \[Delta] Sin[2 f \[Pi] t + \[Phi]] + F2y \[Delta] Sin[2 f \[Pi] t + \[Phi]] == 0 -F1x + 6 a \[Pi] \[Micro] (Vxc + d \[CapitalOmega] Cos[\[Theta]1] + d \[Omega]1z Cos[\[Theta]1] - 2 f \[Pi] \[Delta] Cos[2 f \[Pi] t + \[Phi]]) == 0 -F1y + 6 a \[Pi] \[Micro] (Vyc - L0 \[CapitalOmega] + d \[CapitalOmega] Sin[\[Theta]1] + d \[Omega]1z Sin[\[Theta]1] - \[Delta] \[CapitalOmega] Sin[ 2 f \[Pi] t + \[Phi]]) == 0 -F2x + 6 a \[Pi] \[Micro] (Vxc + d \[CapitalOmega] Cos[\[Theta]2] + d \[Omega]2z Cos[\[Theta]2] + 2 f \[Pi] \[Delta] Cos[2 f \[Pi] t + \[Phi]]) == 0 -F2y + 6 a \[Pi] \[Micro] (Vyc + L0 \[CapitalOmega] + d \[CapitalOmega] Sin[\[Theta]2] + d \[Omega]2z Sin[\[Theta]2] + \[Delta] \[CapitalOmega] Sin[ 2 f \[Pi] t + \[Phi]]) == 0 -T1z + 8 a^3 \[Pi] (\[CapitalOmega] + \[Omega]1z) \[Micro] == 0 -T2z + 8 a^3 \[Pi] (\[CapitalOmega] + \[Omega]2z) \[Micro] == 0 where the variables I want are: {Vxc, Vyc, \[CapitalOmega], F1x, F1y, T1z, F2x, F2y, T2z} And the initial conditions are: \[Omega]1z = \[Omega]; \[Omega]2z = \[Omega]; \[Theta]1 = \[Theta]10 + \[Omega]*t; \[Theta]2 = \[Theta]20 + \[Omega]*t; The specific case that gives independent answers is: \[Omega]1z = \[Omega]; \[Omega]2z = -\[Omega]; \[Theta]1 = -Pi/6 + \[Omega]*t; \[Theta]2 = Pi/6 - \[Omega]*t; I also found out that as long as in the initial conditions the absolute of [Theta]10 and [Theta]20 are the same the answer is independent but as soon as there's a phase difference the answers become dependent. Answer
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Posted 7 months ago
 What exactly are the computations? I see a description of the setup but not what function is being invoked. Solve? Answer
Posted 7 months ago
 Yes, I used both solve and Nsolve. Same error and results. and the error is: Solve::svars: Equations may not give solutions for all "solve" variables. Answer
Posted 7 months ago
 Crossposted here. Answer